What is a solution to the following integral:

$$\int_0^\infty \, \frac{\cos{kt}}{\pi}\,\mathrm{d}k\,?$$

I have tried to evaluate this in the usual way:

$$\begin{align}\frac{1}{\pi}\int_0^\infty\,\cos{kt}\,\mathrm{d}k &= \frac{1}{\pi t} \left[\sin{k t}\right]\bigg|_{k\rightarrow 0}^{k\rightarrow \infty} \\ &=\frac{1}{\pi t} \left[\lim_{k\rightarrow \infty}\sin{k t} - \sin 0\right] \\ &=\frac{1}{\pi t} \lim_{k\rightarrow \infty}\sin{k t},\end{align}$$

but $\lim_{k\rightarrow \infty}\sin{k t}$ doesn't converge within infinity.

What is the trick here? Note that this is a homework question, so I don't need the complete solution, just directions where to start.

  • 1
    $\begingroup$ this integral is divergent in the usual sense. a way to interpret it would require generalized functions or distributions. are you familiar with that? $\endgroup$ – tired Nov 11 '15 at 0:50
  • $\begingroup$ @tired, that rings a bell, yes. Upon reading about them, though, I can't see how it could be applied. $\endgroup$ – Harry Smith Nov 11 '15 at 1:14
  • $\begingroup$ do you know the fouriertransform of $1$? $\endgroup$ – tired Nov 11 '15 at 1:18
  • $\begingroup$ @tired, the delta-function, right? EDIT: I see the answer! Thank you for your pointers! $\endgroup$ – Harry Smith Nov 11 '15 at 1:35
  • $\begingroup$ yes. and u also know the representation of $cos$ through exponential functions? $\endgroup$ – tired Nov 11 '15 at 1:36

To answer your question, there is no finite value for the integral.

As you showed, the integral simplifies to the value:

$$\frac{1}{\pi t} \lim_{k\rightarrow \infty}\sin{k t}$$

Looking at the limit, as $k$ approaches infinity, the function $\sin{(kt)}$ will oscillate between the values $-1$ and $1$, and therefore won't converge to a finite value.

You could use taylor series of $\sin x$ to get a value through approximation, but the integral itself doesn't converge to any value.

  • $\begingroup$ I vaguely remember seeing a theorem that stated something like $\int_0^\infty g(x)\,f(x)\,dx = \sum f(x)/g'(x)$, does this sound familiar? $\endgroup$ – Harry Smith Nov 11 '15 at 1:18
  • $\begingroup$ @HarrySmith I'm not familar with the equation. Do you have a site where that idea can be further explained? $\endgroup$ – Varun Iyer Nov 11 '15 at 1:19
  • $\begingroup$ EDIT: The theorem I was thinking of was that $\int_a^b \, g(x)\:\delta\left(f(x)\right) \mathrm{d}x = \sum_{n=1}^N g(x_n)/\left|f'(x_n)\right|$. $\endgroup$ – Harry Smith Nov 11 '15 at 2:02

The integral doesn't converge to any finite value. A graph of the function you're integrating would be a transformed version of $\cos k$, which oscillates back and forth. The area underneath would also oscillate back and forth, which is what your answer already tells you.


I have found that a solution can can be inferred from equation 2.3.36 (pg. 137) in Dubin's Numerical and Analytical Methods for Scientists and Engineers using Mathematica, which finds that

$$\lim_{k\rightarrow\infty} \frac{\sin{kt}}{\pi t} = \delta(t),$$

meaning that

$$\int_0^\infty \frac{\sin{k t}}{\pi}\,\mathrm{d}k = \delta(t)$$

a result, admittedly, not satisfactory for mathematics stack exchange since the limit is not well defined (and is not finite) but it is noteworthy and useful nonetheless.


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